Sum of focal radii = 2a property: Question bank

Definition Every symbol used on this page (read this first)

Picture a squashed circle — an ellipse — drawn on a chalkboard, wider than it is tall. The figure below labels each name so no symbol is a mystery:

$a$ = the semi-major axis = half the longest width of the oval, measured from the centre to the far tip (a vertex at $(a, 0)$ ). The full longest width is $2 a$ .
$b$ = the semi-minor axis = half the shortest width, centre to the top tip $(0, b)$ . Always $b < a$ here.
The two foci are two special inside points sitting on the long axis: $S (a e, 0)$ (right) and $S^{'} (- a e, 0)$ (left).
$e$ = the eccentricity, a number with $0 < e < 1$ that says how stretched the oval is ( $e = 0$ is a perfect circle). The foci sit at distance $a e$ from the centre.
$x$ = the horizontal coordinate of a point $P (x, y)$ that lies on the ellipse; it always satisfies $∣ x ∣ \leq a$ .
$S P$ = the straight-line distance from the point $P$ to the right focus $S$ . $S^{'} P$ = the distance from $P$ to the left focus $S^{'}$ . These two are the focal radii.

Every reveal is Statement ::: reasoning. The reasoning line always says why, never a bare yes/no.

True or false — justify

For a point on the ellipse,

S P + S^{'} P

can occasionally be less than

2 a

P

is near a vertex.

False — the sum is locked at exactly

2 a

everywhere, including the vertices; at

(a, 0)

it is

(a - e a) + (a + e a) = 2 a

The gardener's string trick works because the string length equals

2 b

False — the string length is the constant sum

S P + S^{'} P = 2 a

(the major axis span), not

2 b

;

b

never appears as a sum of radii.

The sum

S P + S^{'} P

stays constant, and so does the difference

S^{'} P - S P

False — only the sum is constant; the difference is

2 e x

, which changes with the point's

x

-coordinate.

If two ellipses share the same foci but have different shapes, they must have the same

2 a

False — same foci fix

a e

, but

a

(hence

2 a

) can differ; a fatter ellipse simply has a larger string / larger

a

For a circle (an ellipse with

e = 0

), the "sum of focal radii" equals the diameter.

True — with

e = 0

the two foci merge at the centre, so

S P = S^{'} P = a = r

, and their sum is

2 a = 2 r =

diameter.

The formula

S P = a - e x

works for the left focus too, just swap the sign of

x

False — the formula itself differs by focus: right focus gives

a - e x

, left focus gives

a + e x

; you don't re-sign

x

, you use the correct expression.

If the major axis lies along the

y

-axis, the sum of focal radii is

2 b

False — it is always

2 \times (semi-major axis)

; if the vertical axis is longer, call that length

a

, and the sum is still

2 a

, never the smaller semi-axis.

The property

S P + S^{'} P = 2 a

is a theorem you must derive from the equation.

Partly false — it is the defining property (constant-sum locus); the standard equation is derived from it, so you can also treat it as the starting axiom.

Spot the error

"For

\frac{x ^{2}}{16} + \frac{y ^{2}}{25} = 1

, we have

a = 4

so the focal-radii sum is

8

Error: the larger denominator (

25

) is under

y^{2}

, so

a = 5

and the sum is

2 a = 10

;

4

is the semi-minor axis here.

"At the point

(0, b)

the two focal radii are unequal, one big and one small."

Error:

(0, b)

is on the minor axis, equidistant from both foci, so

S P = S^{'} P = a

— they are equal, each exactly half the string.

"Right focus

S (a e, 0)

: as

x

grows the point moves right toward

S

, so

S P

should grow."

Error: moving toward a focus shrinks the distance; that's exactly why

S P = a - e x

decreases as

x

increases.

S P = a - e x

can go negative for large

x

, so the formula breaks."

Error: on the ellipse

x

is capped at

∣ x ∣ \leq a

and

0 < e < 1

, so

a - e x \geq a - e a = a (1 - e) > 0

; it never goes negative, and the formula only claims to hold on the curve anyway.

"Since the sum is

2 a

for the ellipse and the difference is

2 a

for the hyperbola, they're basically the same constant."

Error: the word is the same but the operation differs — sum vs absolute difference — which produces oppositely-curved locus shapes.

"Foci at

(\pm 3, 0)

with

a = 2

gives a valid ellipse with sum

2 a = 4

Error: foci need

a e = 3

with

e < 1

, so

a > 3

; taking

a = 2 < 3

is impossible — the "foci" would lie outside the curve, and no ellipse exists.

"The focal radius via directrix is

S P = e \cdot x

Error: the right focus

S

pairs with the right directrix, the vertical line

x = a / e

; the rule is

S P = e \cdot d

where

d

is the perpendicular (horizontal) distance

d = \frac{a}{e} - x

from

P (x, y)

to that line — not the raw coordinate

x

. Check:

e (\frac{a}{e} - x) = a - e x = S P

. ✓

Why questions

Why does the

x

-term cancel when we add the two focal radii?

Because

S P = a - e x

and

S^{'} P = a + e x

carry equal-and-opposite

e x

terms; moving nearer one focus loses exactly what moving farther from the other gains.

Why must the constant sum be the major axis length and not something smaller?

The stretched string has to reach the far vertices; the maximum span it must accommodate is the full major axis

2 a

, which then holds everywhere.

Why does the derivation force us to substitute

b^{2} = a^{2} (1 - e^{2})

That substitution is precisely what makes

(x - a e)^{2} + y^{2}

collapse into the perfect square

(a - e x)^{2}

, letting the square root simplify cleanly.

Why can we forecast

S^{'} P

from

S P

without knowing the point's coordinates?

Because their sum is a fixed constant

2 a

; once

S P

is given,

S^{'} P = 2 a - S P

instantly, no geometry required.

Why is the sum property called the definition of the ellipse rather than just a fact about it?

Every point with constant focal-radii sum lies on the curve and vice versa, so the constant-sum condition generates the ellipse — it's the locus rule itself.

Why does the difference

S^{'} P - S P = 2 e x

vanish exactly on the minor axis?

On the minor axis

x = 0

, so

2 e x = 0

; the point is symmetric between the foci and both radii are equal.

Edge cases

At the two vertices

(\pm a, 0)

— do the focal radii still add to

2 a

Yes — at

(a, 0)

they are

a - e a

and

a + e a

, summing to

2 a

; the property holds at the extreme points too.

As eccentricity

e \to 0

, what happens to the two focal radii?

Both approach

a

(the foci merge at the centre into a circle), so each radius is the constant radius and the sum

2 a

becomes the diameter.

e \to 1

(foci sliding toward the vertices), does the sum change?

No — the sum stays

2 a

; only the difference

2 e x

grows, so the ellipse looks increasingly stretched while the string length is unchanged.

P

coincides with a focus, is the "point on the ellipse" analysis valid?

No — a focus lies inside the ellipse, not on it, so the focal-radii property (which is about points on the curve) simply doesn't apply there.

For a point at the very end of the minor axis

(0, b)

, what are the two radii individually?

Each equals

a

(the semi-major length), since the point is symmetric between the foci; together they still give

2 a

Can the sum of focal radii ever exceed the major-axis length

2 a

No — it equals exactly

2 a

at every point on the curve; the taut string can never measure more than its own fixed length.